3cot²x - 5cotx = -2
Lets pretend cotx = t
3t² - 5t = -2
3t² - 5t + 2 = 0
factor polynomial
(3t-2)(t-1) = 0
3t-2 = 0
t-1 = 0
then "t" which is also cotx is equal to 1 or 2/3
the answer to this very difficult question, but very interesting at the same time 3 hours
Answer:
16
Step-by-step explanation:
6+2(t) --> 6+2(5) --> 6+10=16
Answer:
m=3/4
Step-by-step explanation:
We know that
tan²(A)+1 = sec²(A)--------> tan²(A)= sec²(A)-1
tan²(A)= (4/3)²-1------> tan²(A)=16/9-1-----> 7/9
tan²(A)=7/9
if the angle A is in the IV quadrant
then
tan (A) is negative
therefore
tan (A)=-√(7/9)-----> tan (A)=-√7/3
so
cot (A)=1/tan(A)------> cot (A)=-3/√7----> -(3/7)*√7
the answers are
a) the identity is tan²(A)= sec²(A)-1
b) tan (A)=-√7/3
